Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.

To find the unpaired electrons, we first need to find the oxidation state of Fe in both complex ions. For \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\): The overall charge of the complex ion is -4, and CN is a monodentate ligand with a charge of -1. Since there are 6 CN molecules, the total charge contributed by the ligands is -6. Thus, the oxidation state of Fe in this complex ion is +2. For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): The overall charge of the complex ion is +2, and water (H2O) is a neutral ligand, not contributing any charge. Therefore, the oxidation state of Fe in this complex ion is also +2.

Fe has an electron configuration of [Ar] 3d6 4s2, which means that in its ground state, it has 6 electrons in the 3d orbital and 2 electrons in the 4s orbital. When Fe has an oxidation state of +2, it loses 2 electrons and its electron configuration becomes [Ar] 3d6 for both complex ions.

Next, we need to determine the number of unpaired electrons for each complex ion. For \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\): The electron configuration of Fe in this complex ion is [Ar] 3d6. CN is a strong field ligand, causing the energy difference between the d-orbitals to be large (greater crystal field splitting). This results in a low-spin complex where all six electrons of the Fe(II) ion will occupy the three lower-energy 3d orbitals, following the Aufbau principle, and pairing up with each other. Therefore, this complex will have 0 unpaired electrons and be diamagnetic. For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): The electron configuration of Fe in this complex ion is also [Ar] 3d6. However, H2O is a weak field ligand, resulting in smaller energy differences between the d-orbitals (lower crystal field splitting). Consequently, the Fe(II) electrons will fill the five 3d orbitals before pairing up, following Hund's rule. This complex will have 4 unpaired electrons and be paramagnetic.

Given our findings, it's clear that \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is more likely to be paramagnetic, as it has 4 unpaired electrons while \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) will be diamagnetic with 0 unpaired electrons.